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Difficulty: MediumOverlapping Sets, Statistics, and Data Distributions

At a technology firm, 6060 software engineers work on Project Alpha, Project Beta, or both. Exactly 4040 engineers work on Project Alpha, and exactly 3535 engineers work on Project Beta. Is the average (arithmetic mean) years of experience of all 6060 engineers combined greater than 77 years?

(1) The average years of experience of the engineers who work ONLY on Project Alpha is 88 years, and the average years of experience of the engineers who work ONLY on Project Beta is 55 years.
(2) The average years of experience of all 4040 engineers on Project Alpha is 8.758.75 years.

  1. A
    Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
  2. B
    Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
  3. BOTH statements TOGETHER are sufficient, but NEITHER statement alone is sufficient.Answer
  4. D
    EACH statement ALONE is sufficient.
  5. E
    Statements (1) and (2) TOGETHER are NOT sufficient.

Answer

Both statements together are sufficient to answer the question definitively with a 'Yes', but neither statement alone is sufficient.
Combining both statements provides the exact sum of values across all three mutually exclusive sub-groups (Only Alpha, Only Beta, and Both), yielding a unique overall average of 7.5 years, which yields a definitive 'Yes' answer.

Step-by-Step Solution

1
Determine the sizes of the non-overlapping and overlapping groups.
Group sizes: Only Alpha = 25, Only Beta = 20, Both = 15.
Using the overlapping sets formula N(A∪B)=N(A)+N(B)−N(A∩B)N(A \cup B) = N(A) + N(B) - N(A \cap B), we get 60=40+35−N(A∩B)60 = 40 + 35 - N(A \cap B), which yields N(A∩B)=15N(A \cap B) = 15. Thus, N(Only Alpha)=40−15=25N(\text{Only Alpha}) = 40 - 15 = 25 and N(Only Beta)=35−15=20N(\text{Only Beta}) = 35 - 15 = 20.
2
Evaluate Statement (1) alone.
Statement (1) is NOT sufficient.
Statement (1) gives SOnly A=25×8=200S_{\text{Only A}} = 25 \times 8 = 200 and SOnly B=20×5=100S_{\text{Only B}} = 20 \times 5 = 100. The sum of experience for the 1515 overlap engineers (SBothS_{\text{Both}}) remains unknown. Total mean =300+SBoth60= \frac{300 + S_{\text{Both}}}{60}. If SBoth=150S_{\text{Both}} = 150 (mean 1010), total mean =7.5>7= 7.5 > 7 (Yes). If SBoth=30S_{\text{Both}} = 30 (mean 22), total mean =5.5≤7= 5.5 \le 7 (No). Hence, Statement (1) alone is insufficient.
3
Evaluate Statement (2) alone.
Statement (2) is NOT sufficient.
Statement (2) gives the total experience of all 4040 engineers on Project Alpha: SAlpha=40×8.75=350S_{\text{Alpha}} = 40 \times 8.75 = 350. This means SOnly A+SBoth=350S_{\text{Only A}} + S_{\text{Both}} = 350. Total mean =350+SOnly B60= \frac{350 + S_{\text{Only B}}}{60}. Since SOnly BS_{\text{Only B}} is unknown, the total mean could be 7.57.5 (if SOnly B=100S_{\text{Only B}} = 100) or 6.176.17 (if SOnly B=20S_{\text{Only B}} = 20). Hence, Statement (2) alone is insufficient.
4
Evaluate Statements (1) and (2) together.
Statements (1) and (2) together are SUFFICIENT.
From Statement (1), SOnly A=200S_{\text{Only A}} = 200 and SOnly B=100S_{\text{Only B}} = 100. From Statement (2), SOnly A+SBoth=350S_{\text{Only A}} + S_{\text{Both}} = 350, which implies 200+SBoth=350  ⟹  SBoth=150200 + S_{\text{Both}} = 350 \implies S_{\text{Both}} = 150. The combined total experience is STotal=200+100+150=450S_{\text{Total}} = 200 + 100 + 150 = 450. The overall mean is 45060=7.5\frac{450}{60} = 7.5 years. Since 7.5>77.5 > 7, we get a definitive 'Yes'.

Key Concept

Data Sufficiency evaluation for combined weighted averages and overlapping set partitions
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Topics

Overlapping Sets, Statistics, and Data Distributions

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